Circuit Length

[Mark.W]

Hi Peeps

I have my 2391 coming up in October and I am busy revising.

When determing the circuit length and VD, my notes say to link L - N at the load side and take the loop resistance at the DB.

The example I have is : Loop Resistance = 0.7 ohms (R) and csa of cable = 1mm2 (S), calculate the circuit length??

The formula they give is Circuit Length = 29.4 * R * S.

My question is where does the 29.4 come from??

Cheers
Mark

 

[malcolmsanford]

I always calculated circuit length by the R1 + R2 using the resistance of the cable given in the OSG guide Table 1A

So, if you divide the result of the R1+R2 test by this value it will give you the length of the cable. So say you have 2.5mm T+E with a 1.5mm CPC then that is 19.51.

You take your R1 + R2 reading and get say 0.54

0.54 ÷ (19.51 ÷ 1000) = 28.42 meters.

Is that 29.4 just a token resistivity example

 

[Mark.W]

Hi mate

The example they give is Circuit Length = 29.4 * R * S
The answer they give is Circuit Length = 29.4 * 0.7 * 1 = 20.6 metres
In other words they're muliplying the Loop Resistance of 0.7 by the csa of the cable 1mm2 by 29.4!!

I have never come accross this equation before? Confused.com

 

[Deleted member 26818]

I think the 29.4 is made up. It's supposed to be the resistance in millohms per meter of the cable.
According to Table A.1 in GN3, it should be 36.2 when measured with an ambient temperature of 20ºC.

 

[topquark]

I think that's just a, poor, example.

Formula should be VD = ((mV/A/m) x L x Ib)/1000)

Where L is in meters and Ib is in Amps with the coefficients taken from appendix 4 of BS7671.

 

[Deleted member 26818]

I think that's just a, poor, example.

Formula should be VD = ((mV/A/m) x L x Ib)/1000)

Where L is in meters and Ib is in Amps with the coefficients taken from appendix 4 of BS7671.

Aren't those values in Appendix 4, for when the conductors are at operating temperature?

 

[topquark]

Yes, spin, there are correction factors that can be applied as necessary if it's not at 20degC

 

[malcolmsanford]

Sorry TQ I thought he was trying to determine the Length of the cable

 

[Deleted member 26818]

I like to use the values in Table A.1 in GN3, as you don't have to apply correction factors.

 

[topquark]

Ah yes, I see what you mean, yes, just grabbed my BGB and the tables are listed as for operating temperature, oops

 

[topquark]

Sorry TQ I thought he was trying to determine the Length of the cable

I think it's me that got it wrong here, surprise surprise read the question a bit tooooo quick.

 

[andyb]

Hi Peeps

I have my 2391 coming up in October and I am busy revising.

When determing the circuit length and VD, my notes say to link L - N at the load side and take the loop resistance at the DB.

The example I have is : Loop Resistance = 0.7 ohms (R) and csa of cable = 1mm2 (S), calculate the circuit length??

The formula they give is Circuit Length = 29.4 * R * S.

My question is where does the 29.4 come from??

Cheers
Mark

If you've measured the l/n resistance, why do you need to work out the circuit length?

V=IR so you have R you just need the design current.

So with a L/N loop of .7 as per your example, if the design current was say 6 amps the volt drop would be .7X6 =4.2 volts. Job done.

 

[JUD]

If you've measured the l/n resistance, why do you need to work out the circuit length?

V=IR so you have R you just need the design current.

So with a L/N loop of .7 as per your example, if the design current was say 6 amps the volt drop would be .7X6 =4.2 volts. Job done.

That would actually be 0.7 x 6 x 1.2 = 5V as the resistance was measured at 20°C and volt drop needs to be corrected for the cable operating temperature of 70°C.

 

[andyb]

That would actually be 0.7 x 6 x 1.2 = 5V as the resistance was measured at 20°C and volt drop needs to be corrected for the cable operating temperature of 70°C.

You're right Jud, I'm just making the point that if the cable is already installed you are just verifying the volt drop, you only need to calculate the vd when you are designing a circuit.
Or is this something required for the 2391 exam? I must admit I never came across it.

 

[JUD]

Maybe just to show that you know how to work out the length of a circuit from resistance.

I must admit I've never done it using the formula in the OP. I do it the same way as malcolm said in post #2.

So it would be...

L = R ÷ (mΩ/m ÷ 1000)
L = 0.7 ÷ (36.2 ÷ 1000)
L = 0.7 ÷ 0.0362
L = 19.33m

L = length of circuit, R = measured resistance, mΩ/m = resistance per meter of 2 x 1mm² copper conductors (from Table 9A OSG or Table A.1 GN3)

 

[Deleted member 26818]

I take the measured value of resistance (0.7Ω), divide it by the value given in Table A.1 in GN3 or Table 9A in the OSG (36.2mΩ), and then multiply it by 1000, to change mΩ to Ω.
0.7 ÷ 36.2 = 0.01933 x 1000 = 19.33.

 

[telectrix]

or you could just use a tape measure. only thing is i don't know how to correct the tape for expansion when warm.

 

[Richard Burns]

I have been trying to work this out and, as in the case of others, have not seen this equation before. However the calculation does work out but the number 29.4 should be 27.6 by my calculations.

The proportional constant (in this case given as 29.4)

For Length L, resistance R and resistance per meter Rm
L = R / Rm
Or L = R * 1/Rm

When measuring at the db with L and N connected at the load you are measuring the resistance of twice the length of the circuit; once for the line conductor and then back for the neutral conductor.

So L = (R/2) * 1/Rm
This can be rearranged to give
L= R* 1/(2*Rm)


Given the resistance per meter of 1mm2 copper at 18.1mohm/m= 0.0181ohm/m
(From OSG table 9A appx 9 for a single 1mm2 cable (not the R1+R2 value)

1/(2*Rm) = 1/ 2*0.0181 = 27.6

So for a 1mm2 cable the length = R * 27.6

The length of the cable for any given resistance will be proportional to the csa.
I.e if the resistance R for a 1mm2 copper conductor gives a cable length of 1000m
Given the same resistance R but for a 2.5mm2 copper conductor the length must be increased by 2.5 i.e. 2500m

Therefore for any size cable the length L in relation to the resistance measured R and the csa S is as follows.

L = R * S * 27.6

Unfortunately this does not answer the original question where did 29.4 come from!
But at least it is closer

 

[Deleted member 26818]

Yes, I got 27.6 as well.
It would be nice if there were a table of values, that you could then using a simple calculation X multiplied by the measured resistance = the length.
I generally end up doing the calculation two, three or how ever many times, just to confirm my results.
It doesn't help if you hit the + key on the calculator when you should have hit the x key, or you miss off one of the zeros when dividing by 1000.
Then there's those bloody scientific calculators!

 

[topquark]

or you could just use a tape measure. only thing is i don't know how to correct the tape for expansion when warm.

Using the coefficient of linear expansion of steel (assuming it's a steel tape)